# Suppose you ask a friend to randomly choose an integer between 1 and 10, inclusive. What is the probability that the number will be more than 4 or odd

Suppose you ask a friend to randomly choose an integer between 1 and 10, inclusive. What is the probability that the number will be more than 4 or odd? (Enter your probability as a fraction.)
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Brittany Patton

6 of the 10 integers between 1 and 10 inclusive are more than 4 (that is, 5,6,7,8,9,10).
The probability is the number of favorable outcomes divided by the number of possible outcomes: $P\left(>4\right)=\mathrm{#}$of favorable outcomes $\mathrm{#}$ of possible outcomes=$\frac{6}{10}$
5 of the 10 integers between 1 and 10 inclusive are odd (that is, 1,3,5,7,9). P(odd)=# of favorable outcomes/# of possible outcomes=$\frac{5}{10}$
3 of the 10 integers between 1 and 10 inclusive are move than 4 and odd (that is, 5,7,9). $P\left(>4$ and odd$\right)=\mathrm{#}$ of favorable outcomes/# of possible outcomes=$\frac{3}{10}$
Use the General addition rule for any two events: $P\left(AUB\right)=P\left(A\right)+P\left(B\right)-P\left(A\bigcap B\right)$
P(>4 or odd) $=P\left(>4\right)+P\left(odd\right)-P\left(>4\text{and odd}\right)=\frac{6}{10}+\frac{5}{10}-\frac{3}{10}=\frac{6+5-3}{10}=\frac{8}{10}=\frac{4}{5}$